Question

5. (60 points) Water is the working fluid in a regenerative Rankine cycle with one closed feedwater heater. Superheated vapor enters the first turbine at 15 MPa and 550°C. The cycle has a closed feedwater heater using extracted steam at 1 MPa. Condensate drains from the feedwater heater as saturated liquid at 1MPa and is trapped into the condenser. The feedwater leaves the heater at 15 MPa and a Temperature equal to the saturation temperature at IMPa. The mass flow rate of the steam is 120 kg/s. The sentropic efficiencies of the turbines and the pump ore 85% 4. 173.88 kg, h 7628 kJ/kg. Show all work and indicate any tables used Credit will not be given if sufficient work is not present. 050 State (10 points) Find he and he 5. .2005 AWS 6.52-0.594 260 5052 9.22 .594 SS SS S S -9 -623 - 2.1362 6.98 Ehsen 5 .573- 13.04.2016 h . 2642.195 205.39 € 21.1223 (15 points) Find hot and 2 1 + 1.014 . ) Stut . 1995 ha=761 suala nga 262.048/6) shesheeprom hos 762.51 mily 10 . (10 points) Find the mass percentage y leaving the first turbine. Yo hacks 362.01 - 145.95 Wi-hg 29429 - 32.91 0.297

Answer 1

Given b Pa= IMPE P=15 MPa Z hi=8450:4 kJ/kg / Table A-6) Ti=550°C J 6. = 6:523 kJ/kg-k P2 = 1 MPa 2 has=2748.97 kJ/kg (table A-5) si - Sas J 229 = 0.986 P 77,-0-85-hi-ha - 3450:4-he Michas 3450-4-2748.94 12=2854.18 kJ/kg 7 Tac 211.3°C (Table A-6) H2=285418 kJ/ka Jos2= 6:4416 kJ/kg-k P3=8KPR 2 has < 2110.87 kJ/kg (Table A-5) S3=52 JX35=0.806 77 -0.85 = Ha-ha -2854.18-13 ha-h35 2854.18-2110.87 ho=2222:36 KJ/kg PA=8 KPR 224=14=143.88 KJlka (Table A-5) 24 = 0 JV4=V4 -0.0010084 mol kg 455 =h4+ 14 (Po-P4) = 178.88 +0.0010084 (100 455=174.88 kJ/kg lp=0.85 = 155-h4 = 174.88-193.88 25-14 15-173.88 15 = 175.05 KJ/kg b

PE = 15 MPa = 470.5 kJ/kg (Table A-7) TG = Tsat@IMPa = 149.9°C J Pe=IMPR Z hy = 762.81 kJ/kg 27= OJ hochg=762.81 RTRS Applying mass and energy balance to closed feedwater heater y=46-45 = 7705-175-05 ha-h7 2854,18-462-81 y = 0.2841 Wit = michi-h2)+(1-4) (14-ha) = 120|(3450.4-2854.18)+(1-0.2847)(192222-36-173-88) Wt=247,349 KLU

wpMoha=(17505-173-88]120 wp=140:4 KW Net power developed, linet = hit-wp = 247,379-140:4 Kinet- 247,238 KW winet =2471238 MW Rate of heat transfer through boiler dinamchi-h6) =120 (3450-4-770.5)* la sin= 321.588 MW Thermal efficiency is 7th = wnet = 241238 321.588 Qin 7th =76'8%